Radical Functions

Intuition

Squaring a number is easy: feed in $5$ and out comes $25$ . A radical function runs that machine backward. It starts from the area of a square and hands you back the side, starting from $25$ and returning $5$ . You first met the square root as an operation, back when you learned exponents and roots. Here the square root becomes a function in its own right, with an input, an output, a domain, and a graph. Because a radical undoes a power, everything about it flows from a single question: which number, raised to this power, gives what I started with? Answering that carefully is what explains why square roots refuse negative inputs, and why the graph starts at a corner and bends the way it does. Answering that question also explains why solving a radical equation can hand you an answer that turns out to be false.

Learning goals

Solve radicand $\ge 0$ for an even-index domain
Accept every real input for an odd-index root
Draw $y = \sqrt{x}$ from its corner, rising ever slower
Place the corner at $(h, k)$ in $y = a\sqrt{x - h} + k$
Check every candidate, since squaring adds extraneous ones

What a radical function is

Squaring sends $3$ to $9$ and $7$ to $49$ . The square root asks the reverse question: which nonnegative number, squared, gives the input? Packaging that question as a function gives the square-root function

f(x) = \sqrt{x}.

By agreement $\sqrt{x}$ means the principal root, the one that is not negative. Both $4$ and $-4$ square to $16$ , but $\sqrt{16}$ is defined to be $4$ alone. That definition makes $f$ return exactly one output per input, so $f$ is a genuine function.

The same idea works for any power. The cube root $\sqrt[3]{x}$ undoes cubing, and in general the $n$ th root $\sqrt[n]{x}$ undoes raising to the $n$ th power. The little number $n$ is the index of the radical; a square root is the case $n = 2$ , and its index is left unwritten. Each of these is the same thing you already know from rational exponents, since taking an $n$ th root is raising to the power $\tfrac{1}{n}$ :

\sqrt{x} = x^{1/2}, \qquad \sqrt[3]{x} = x^{1/3}, \qquad \sqrt[n]{x} = x^{1/n}.

A radical function is any function whose rule contains the variable under a root like this. The rest of the lesson works out the two features that make radicals behave unlike the polynomials of earlier chapters: a restricted domain, and a graph that bends.

Domain: which inputs a root will accept

A polynomial such as $x^2 - 3x$ accepts every real number. A radical is fussier. Try to compute $\sqrt{-9}$ and you are asking for a number whose square is $-9$ , and no real number squares to a negative. So $\sqrt{-9}$ has no real value, and $-9$ cannot be in the domain of $f(x) = \sqrt{x}$ . The same objection rules out every negative input.

Why an even-index root rejects negative inputs#

Start from the definition. The value $\sqrt{a}$ is the nonnegative number whose square is $a$ , so if $\sqrt{a} = b$ then $b$ must satisfy $b^2 = a$ . The question is whether any real number $b$ can have a negative square. A positive times a positive is positive, a negative times a negative is again positive, and $0^2 = 0$ , so for every real number $b$ ,

b^2 \ge 0.

A square is never negative. If $a$ is negative, there is no real $b$ with $b^2 = a$ , so $\sqrt{a}$ has no real value. That is exactly why the domain of $f(x) = \sqrt{x}$ is every $x$ with $x \ge 0$ . The same reasoning applies to any even index: the input under a fourth root, a sixth root, and so on must be zero or greater.

Odd-index roots escape the restriction, and the sign test shows why. A cube can be negative, since $(-2)^3 = -8$ , which makes $\sqrt[3]{-8} = -2$ a perfectly good real number. As $b$ runs through every real number, $b^3$ runs through every real output exactly once, so $\sqrt[3]{x}$ is defined for all real $x$ . The domain shrinks only when the index is even, because only an even power is forced to stay nonnegative.

To find the domain of a more involved radical, you do not need a new rule. For an even root, whatever sits under the radical must be greater than or equal to zero. So for an even root, set the radicand (the expression under the root) to be nonnegative and solve the resulting inequality.

Worked example 1 Find the domain of three radical functions

Handle each root by looking only at what it demands of the radicand.

For $f(x) = \sqrt{2x - 6}$ , the index is even, so the radicand cannot be negative. Require it to be nonnegative and solve:

2x - 6 \ge 0 \quad \Longrightarrow \quad x \ge 3.

The domain is all $x \ge 3$ .

For $g(x) = \sqrt{5 - x}$ , again set the radicand to be nonnegative, and remember that dividing by a negative flips the inequality:

5 - x \ge 0 \quad \Longrightarrow \quad -x \ge -5 \quad \Longrightarrow \quad x \le 5.

The domain is all $x \le 5$ . Notice the domain runs to the left here, because raising $x$ makes the radicand smaller.

For $h(x) = \sqrt[3]{x + 4}$ , the index is odd, so the cube root accepts every real number no matter the sign of the radicand. The domain is all real numbers, with nothing to exclude.

Check your understanding

What is the domain of $f(x) = \sqrt{12 - 3x}$ ?

x \le 4

x \ge 4

x \le -4

D all real numbers

The graph of the square root

To plot $y = \sqrt{x}$ , choose inputs whose roots come out whole, the perfect squares, so the points are easy to place:

$x$	$0$	$1$	$4$	$9$	$16$
$y = \sqrt{x}$	$0$	$1$	$2$	$3$	$4$

The curve begins at the corner $(0, 0)$ , since $\sqrt{0} = 0$ , and it climbs to the right forever. But the climb keeps slowing: getting from output $1$ to output $2$ costs three units of input (from $1$ to $4$ ). Getting from output $2$ to output $3$ then costs five units of input (from $4$ to $9$ ). Each additional unit of height demands a wider stretch of input, so the graph rises steeply near the start and then flattens.

That shape is not arbitrary. Saying $y = \sqrt{x}$ says two things at once: $y \ge 0$ (the root is never negative) and $y^2 = x$ (squaring the output returns the input). Read on its own, the equation $y^2 = x$ describes a parabola lying on its side, opening to the right, because for each height $y$ the input is $x = y^2$ . Keeping only the part with $y \ge 0$ leaves the upper half of that sideways parabola, and that half is exactly the square-root graph.

There is a second way to see the same curve. Since the square root undoes squaring on nonnegative inputs, $y = \sqrt{x}$ is the inverse of $y = x^2$ restricted to $x \ge 0$ . So the graph of $y = \sqrt{x}$ is the reflection of that parabola across the line $y = x$ , the mirror you met when studying inverse functions. Reflecting swaps each point $(a, b)$ into $(b, a)$ , which turns the steep, upward half-parabola into the flat, rightward square-root curve.

The square-root curve is the squaring curve seen in a mirror. The dashed curve is y = x squared for x at least 0. Reflecting it across the dashed line y = x swaps every point (a, b) into (b, a) and produces the solid curve y = the square root of x. The point (2, 4) on the parabola becomes (4, 2) on the root, and both curves meet on the mirror at (0, 0) and (1, 1).

The graph also settles the range. The outputs of $\sqrt{x}$ are the heights the curve reaches, and starting at $0$ it climbs through every positive value without bound, so the range is every $y \ge 0$ . Every nonnegative number is hit exactly once, since a target height $y$ is the root of $x = y^2$ .

Reading transformations off the equation

You do not have to build a fresh table for every radical function. Each one is a shifted and scaled copy of $y = \sqrt{x}$ . The moves behind that copy are the same shifts, stretches, and reflections from the graphing lessons, now applied to the root. The general form is

y = a\sqrt{x - h} + k.

Read the three pieces separately, exactly as you did for parabolas:

The $x - h$ inside the root shifts the graph right by $h$ . It also sets the domain, because the radicand $x - h$ must be nonnegative, giving $x \ge h$ .
The $+k$ outside shifts the graph up by $k$ .
The $a$ outside stretches the graph vertically by $\lvert a \rvert$ , and when $a$ is negative it also reflects the graph across the x-axis, flipping the curve downward.

Putting these together, the corner that used to sit at $(0, 0)$ now sits at $(h, k)$ . The domain is $x \ge h$ , and the range is $y \ge k$ when $a > 0$ or $y \le k$ when $a < 0$ . The range splits into two cases because the curve leaves the corner heading up or down according to the sign of $a$ .

Shifting the corner. The dashed curve is y = the square root of x, starting at (0, 0). The rule y = the square root of (x minus 2), plus 1 moves the whole curve right 2 and up 1, so the solid curve starts at the corner (2, 1). The shape is identical; only the starting corner has moved, which is why the domain becomes x at least 2 and the range becomes y at least 1.

Worked example 2 Graph $y = \sqrt{x - 2} + 1$ and state its domain and range

Match the rule to the form $y = a\sqrt{x - h} + k$ . Here $a = 1$ , $h = 2$ , and $k = 1$ , so nothing is stretched or flipped, and the whole curve is only shifted.

The inside $x - 2$ moves the graph right by $2$ , and the $+1$ moves it up by $1$ , so the corner travels from $(0, 0)$ to

(h, k) = (2, 1).

From there the curve has the ordinary square-root shape. To place a second point, pick an input that makes the radicand a perfect square, say $x = 6$ : then $\sqrt{6 - 2} + 1 = \sqrt{4} + 1 = 3$ , giving the point $(6, 3)$ .

The domain comes from the radicand: $x - 2 \ge 0$ , so $x \ge 2$ , which matches the corner starting at $x = 2$ . Since $a = 1 > 0$ the curve rises from the corner, so the outputs start at $1$ and grow, and the range is $y \ge 1$ .

Worked example 3 Describe the graph of $y = -2\sqrt{x}$

Read $y = -2\sqrt{x}$ as $y = a\sqrt{x}$ with $a = -2$ , a single coefficient doing two jobs at once.

Its size, $\lvert a \rvert = 2$ , is a vertical stretch: every output of $\sqrt{x}$ is doubled. Its sign, negative, reflects the graph across the x-axis, turning the curve downward. Tracking the anchor points of $y = \sqrt{x}$ through the rule makes both effects concrete:

(1, 1) \longrightarrow (1, -2), \qquad (4, 2) \longrightarrow (4, -4).

The corner $(0, 0)$ stays fixed, since $-2 \times 0 = 0$ . The domain is unchanged at $x \ge 0$ , because the radicand $x$ is untouched. But the outputs are now zero or negative, so the range flips to $y \le 0$ . The curve now drops away from the origin into the fourth quadrant instead of climbing into the first.

Check your understanding

The graph of $y = \sqrt{x + 3} - 5$ is the graph of $y = \sqrt{x}$ shifted how, and where is its corner?

A left

3

, down

5

; corner

(-3, -5)

B right

3

, down

5

; corner

(3, -5)

C left

3

, up

5

; corner

(-3, 5)

D left

5

, down

3

; corner

(-5, -3)

The three controls below are the three letters of $y = a\sqrt{x - h} + k$ , and the figure restates the domain in words underneath as you move them. The checkpoint you just answered is a state you can reach. Step the inside shift down to $-3$ and the outside shift down to $-5$ , and the corner lands where you said it would.

Watch the domain and the range behave differently, because that difference is the point of this section. The inside shift drags the domain with it: the curve simply does not exist to the left of the corner. So moving $h$ moves the edge of the world, and the sentence under the figure moves with it. The outside shift does not touch the domain at all, it only raises the whole picture, and the range follows it. Then return both shifts to $0$ and set $a$ to $-2$ , which is worked example 3 exactly: the range turns over while the domain sits where it was. Take $a$ all the way to $0$ and the curve flattens into a horizontal line, and note that it is still a line starting at $x = h$ . It still has nothing to its left: the restriction was never about the shape, it came from inside the root.

The domain and the range of $y = a\sqrt{x - h} + k$

Stretch a Inside h Outside k

y = √x. Its start point sits at (0, 0), not shifted at all. From there it rises to the right, at the natural rate, flattening as it goes. It exists only for x of at least 0, because a square root needs the inside to be at least 0.

A square root curve on a coordinate plane, starting at a corner you can move. The inside shift carries the corner sideways and takes the left edge of the domain with it, the outside shift raises the corner and the range with it, and a negative coefficient turns the curve downward without touching the domain.

Solving radical equations

A radical equation has the variable under a root, as in $\sqrt{2x - 1} = 5$ . The plan is to free the variable from the root, and the tool is the inverse relationship you have leaned on all lesson: squaring undoes a square root. Isolate the radical on one side, then raise both sides to the power that matches the index. For a square root you square; for a cube root you cube.

\sqrt{2x - 1} = 5 \quad \Longrightarrow \quad 2x - 1 = 25 \quad \Longrightarrow \quad x = 13.

Squaring turned the awkward equation into a linear one. There is a catch, though, and it is not optional bookkeeping. Raising both sides to a power can manufacture solutions that solve the new equation but not the original. These are called extraneous solutions, and the next argument explains exactly where they come from.

Why raising both sides to a power can create extraneous solutions#

Squaring both sides of an equation is a one-way street. If two quantities are equal, then their squares are equal too: from $A = B$ it always follows that $A^2 = B^2$ . So every genuine solution of the original equation survives the squaring, and none is lost.

The reverse direction is where trouble enters. Knowing $A^2 = B^2$ does not force $A = B$ , because a square erases a sign. The statement $A^2 = B^2$ means

A = B \quad \text{or} \quad A = -B,

since $A$ and $-A$ share the same square. Squaring folds those two possibilities into one, so the squared equation quietly carries the solutions of $A = B$ together with the solutions of $A = -B$ . Any candidate that belongs only to the second family satisfies the squared equation while failing the one you started with.

Those intruders are the extraneous solutions. They are not errors in the algebra; they are the price of an irreversible step, because squaring cannot tell $A$ from $-A$ . The only way to know which candidates are real solutions is to substitute each one back into the original equation and keep those that make it true.

Worked example 4 Solve $1 + \sqrt{2x + 3} = 6$

Isolate the radical before doing anything else. Subtract $1$ from both sides so the root stands alone:

\sqrt{2x + 3} = 5.

Now the two sides are single quantities, so square both to remove the root:

2x + 3 = 25 \quad \Longrightarrow \quad 2x = 22 \quad \Longrightarrow \quad x = 11.

Check the candidate in the original equation, not the squared one:

1 + \sqrt{2(11) + 3} = 1 + \sqrt{25} = 1 + 5 = 6. \checkmark

The check holds, so $x = 11$ is the solution. Isolating first mattered: squaring the equation while the $1$ was still attached would have left a radical behind.

Worked example 5 Solve $\sqrt{x} = x - 2$ and watch an extraneous solution appear

The radical is already isolated, so square both sides. The right side is a binomial, so expand it carefully:

x = (x - 2)^2 = x^2 - 4x + 4.

Move everything to one side to get a quadratic, then factor:

x^2 - 5x + 4 = 0 \quad \Longrightarrow \quad (x - 1)(x - 4) = 0,

so the candidates are $x = 1$ and $x = 4$ . Squaring may have added a stranger, so test both in the original equation $\sqrt{x} = x - 2$ .

For $x = 4$ : the left side is $\sqrt{4} = 2$ and the right side is $4 - 2 = 2$ . The two agree, so $x = 4$ is genuine.

For $x = 1$ : the left side is $\sqrt{1} = 1$ and the right side is $1 - 2 = -1$ . Since $1 \ne -1$ , the candidate $x = 1$ fails, so it is extraneous. It is exactly the intruder the proof predicted: it solves the other branch $\sqrt{x} = -(x - 2)$ , because $1 = -(1 - 2) = 1$ is true. The only solution is $x = 4$ .

Worked example 6 Solve $\sqrt{7x - 5} = x + 1$

Square both sides, expanding the right side as a binomial:

7x - 5 = (x + 1)^2 = x^2 + 2x + 1.

Collect into a quadratic and factor:

x^2 - 5x + 6 = 0 \quad \Longrightarrow \quad (x - 2)(x - 3) = 0,

giving candidates $x = 2$ and $x = 3$ . Test each in the original equation, since squaring might have introduced a false root.

For $x = 2$ : $\sqrt{7(2) - 5} = \sqrt{9} = 3$ , and $x + 1 = 3$ . They match.

For $x = 3$ : $\sqrt{7(3) - 5} = \sqrt{16} = 4$ , and $x + 1 = 4$ . They match too.

This time both candidates check out, so the equation has two solutions, $x = 2$ and $x = 3$ . The lesson is not that the smaller root is always extraneous; it is that you cannot know until you test, so check every candidate in the original equation.

Check your understanding

How many real solutions does $\sqrt{x} = x - 6$ have?

A one B two C none D infinitely many

Common mistakes

Thinking a square root can be negative. The symbol $\sqrt{x}$ means the principal, nonnegative root, so $\sqrt{16} = 4$ , not $\pm 4$ . The two square roots of $16$ are $4$ and $-4$ , but the function returns only the nonnegative one.
Forgetting to restrict the domain. For an even root the radicand must be $\ge 0$ . Before doing anything with $\sqrt{2x - 6}$ , note that $x \ge 3$ ; inputs below that are simply not allowed.
Flipping the inequality the wrong way, or not at all. Solving $5 - x \ge 0$ gives $x \le 5$ , not $x \ge 5$ . When the variable has a negative coefficient, isolating it reverses the inequality.
Squaring term by term. $(x - 2)^2$ is $x^2 - 4x + 4$ , not $x^2 + 4$ . A binomial must be expanded in full; you cannot square each piece separately.
Skipping the check. Raising both sides to a power can create extraneous solutions, so every candidate must be tested in the original equation. Never report a root you have not verified.
Assuming the smaller candidate is the extraneous one. Sometimes both candidates are valid, and sometimes the larger one fails. Only substitution decides which roots are real.

Key takeaways

A radical function puts the variable under a root, as in $f(x) = \sqrt{x}$ or $\sqrt[n]{x} = x^{1/n}$ ; the square root returns the principal (nonnegative) value.
Domain: an even-index radicand must be $\ge 0$ (because a real square is never negative), so solve radicand $\ge 0$ ; an odd-index root accepts every real number.
The graph of $y = \sqrt{x}$ starts at the corner $(0, 0)$ and rises ever more slowly. The same curve is the upper half of the sideways parabola $y^2 = x$ , equivalently the reflection of $y = x^2$ (for $x \ge 0$ ) across $y = x$ . Its range is $y \ge 0$ .
Transformations: in $y = a\sqrt{x - h} + k$ the corner sits at $(h, k)$ and the domain is $x \ge h$ . The factor $a$ stretches by $\lvert a \rvert$ (and reflects across the x-axis when $a < 0$ ), and the range is $y \ge k$ if $a > 0$ or $y \le k$ if $a < 0$ .
Solving: isolate the radical, then raise both sides to the matching power. Squaring is irreversible, so it can add extraneous solutions; check every candidate in the original equation and keep only those that make it true.

Radical Functions

Learning goals

What a radical function is

Domain: which inputs a root will accept

The graph of the square root

Reading transformations off the equation

Solving radical equations

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)